Timeline for Under what conditions can an orientable Riemannian 3-manifold be defined implicitly?
Current License: CC BY-SA 4.0
16 events
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Aug 18, 2022 at 16:45 | answer | added | Ian Agol | timeline score: 4 | |
Jul 20, 2022 at 9:14 | vote | accept | dennis | ||
Jul 19, 2022 at 15:23 | comment | added | Ryan Budney | You want to smooth $f^2$, that way you do not need to modify the function near $\Sigma$. | |
Jul 19, 2022 at 10:44 | comment | added | dennis | @RyanBudney If the smoothed function $\tilde{f}$ is used, then $f(x)$ and $\tilde f(x)$ are not necessarily equal, so I don't think $\Sigma=\tilde f^{-1}(0)$. | |
Jul 16, 2022 at 15:40 | comment | added | Moishe Kohan | For incomplete Riemannian manifolds the answer is negative. | |
Jul 16, 2022 at 8:36 | comment | added | Ryan Budney | The function $f^2$ is smooth near the manifold $\Sigma$, but it could potentially have some non-smooth points far from $\Sigma$. These could be smoothed-away using bump functions. | |
Jul 16, 2022 at 0:55 | comment | added | Ryan Budney | Thanks, ThiKu has answered your question provided the Riemann manifold is complete. | |
Jul 15, 2022 at 23:50 | history | edited | dennis | CC BY-SA 4.0 |
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Jul 15, 2022 at 21:58 | vote | accept | dennis | ||
Jul 15, 2022 at 23:49 | |||||
Jul 15, 2022 at 4:45 | comment | added | Moishe Kohan | In your setting, one can make the submanifold to consist of regular points of the defining function (since orientable 3-manifolds have trivial tangent bundle). | |
Jul 14, 2022 at 23:10 | comment | added | Ryan Budney | Using the word "implicitly" are you referring to the implicit function theorem, i.e. do you want $0$ to be a regular value of $f$? | |
Jul 14, 2022 at 23:04 | answer | added | ThiKu | timeline score: 10 | |
Jul 14, 2022 at 22:51 | history | edited | dennis | CC BY-SA 4.0 |
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Jul 14, 2022 at 20:54 | history | edited | dennis |
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S Jul 14, 2022 at 20:48 | review | First questions | |||
Jul 14, 2022 at 21:18 | |||||
S Jul 14, 2022 at 20:48 | history | asked | dennis | CC BY-SA 4.0 |